Answer:
The answer is "effective stress at point B is 7382 ksi "
Explanation:
Calculating the value of Compressive Axial Stress:
[tex]\to \sigma y =\frac{F}{A} = \frac{4 F}{( p d ^2 )} = \frac{(4 x ( - 40000 \ lbf))}{[ p \times (1 \ in)^2 ]} = - 50.9 \ ksi \\[/tex]
Calculating Shear Transverse:
[tex]\to \frac{4v}{ 3 A} = \frac{4 (75 \ lbf + 25 \ lbf)}{ \frac{3 ( lni)^2}{4}}[/tex]
[tex]= \frac{4 (100 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\ = \frac{400 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\= 0.17 \ ksi[/tex]
[tex]= R \times 200 \ in - P \times 100 \ in = 12500 \ lbf \times\ in[/tex]
[tex]\to \sigma' =[ s y^2 +3( t \times y^2 + t yz^2 )] \times \frac{1}{2}\\\\[/tex]
[tex]= [ (-50.9)^2 +3((63.7)^2 +(0.17)^2 )] \times \frac{1}{2}\\\\=[2590.81+ 3(4057.69)+0.0289]\times \frac{1}{2}\\\\=[2590.81+12,173.07+0.0289] \times \frac{1}{2}\\\\=14763.9089\times \frac{1}{2}\\\\ = 7381.95445 \ ksi\\\\ = 7382 \ ksi[/tex]
Two particles are separated by 0.38 m and have charges of -6.25 x 10-°C and 2.91 x 10-°C. Use Coulomb's law to predict the force between the particles if the distance is cut in half. The equation for Coulomb's law is F = kqi 42, and the constant, k, equals 9.00 x 109 Nm2/C2 2
Answer:
-4.35 × 10^-6 N
Explanation:
i just answered it on ap3x :)
A racing car starts from rest at t=0 and reaches a final speed v at time t. If the acceleration of the car is constant during this time, which of the following statements are true? a) the average speed of the car is v/2; b) the car travels a distance vt; c) the magnitude of the acceleration of the car is v/t; d) the velocity of the car remains constant; or e) none of the statements is true
Answer:
All true statements are shown: a) The average speed of the car is [tex]\frac{v}{2}[/tex], c)The magnitude of the acceleration of the car is [tex]\frac{v}{t}[/tex].
Explanation:
Let prove the validity of each statement:
a) The average speed of the car is [tex]\frac{v}{2}[/tex].
The average speed ([tex]\bar v[/tex]) is defined by the following formula:
[tex]\bar v = \frac{v_{o}+v_{f}}{2}[/tex] (1)
Where:
[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speeds of the racing car.
If we know that [tex]v_{o} = 0[/tex] and [tex]v_{f} = v[/tex], then the average speed of the racing car:
[tex]\bar v = \frac{0+v}{2}[/tex]
[tex]\bar v = \frac{v}{2}[/tex]
The statement is true.
b) The car travels a distance [tex]v\cdot t[/tex].
Since the racing car is accelerating uniformly, the distance travelled by the car is represented by the following kinematic formula:
[tex]x - x_{o}=v_{o}\cdot t + \frac{1}{2}\cdot a\cdot t^{2}[/tex] (2)
Where [tex]a[/tex] is the acceleration of the racing car, measured in meters per square second.
The statement is false.
c) The magnitude of the acceleration of the car is [tex]\frac{v}{t}[/tex].
Since the racing car is accelerating uniformly, the velocity of the racing car is represented by the following kinematic formula:
[tex]v_{f} = v_{o}+a\cdot t[/tex] (3)
Then, we clear the acceleration of the expression:
[tex]a = \frac{v_{f}-v_{o}}{t}[/tex]
If we know that [tex]v_{o} = 0[/tex] and [tex]v_{f} = v[/tex], then the acceleration of the car is:
[tex]a = \frac{v-0}{t}[/tex]
[tex]a = \frac{v}{t}[/tex]
The statement is true.
d) The velocity of the car remains constant.
Since the car accelerates uniformly, the vehicle does not travel at constant velocity.
The statement is false.
A solid disk and a hoop are simultaneously released from rest at the top of an incline and roll down without slipping. Which object reaches the bottom first?
A) The disk arrives first.
B) The hoop arrives first.
C) The one that has the largest radius arrives first.
D) The hoop and the disk arrive at the same time.
E) The one that has the largest mass arrives first.
Answer:
A) The disk arrives first
Explanation:
The disk arrives first, and the reason behind this is that the disk posses smaller moment of inertia, picking up quickly of angular speed of an object for a particular torgue can be determined by ts moment of inertia . So small quantity of energy will be used up as rotational energy, and with the rotational energy the disk will move faster that the hoop.
The option that illustrates the object that reaches the bottom first is A. The disk arrives first.
From the information given, we're told that a solid disk and a hoop are simultaneously released from rest at the top of an incline and rolled down without slipping.
It should be noted that the disk will arrive first due to the fact that it has smaller inertia. In such a case, the disk will arrive first while the hoop arrives later.
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What is force? Define with an example.
Explanation:
A force involves an interaction between two or more objects, and it causes a push or pull between the objects. An example of opposing force include drag due to interaction with an air mass and the force due to friction between two objects.
hope it helps!
A scientist conducted research and plotted his results in the histogram shown below. Which of the
following is the best measure of the center of the data?
Median
Mode
Mean
They are all appropriate
Answer:
Median
Explanation:
we commonly use mean as the measure of the center of the data.But Median is better measure than mean.Extra information:-
Median formula:-
[tex]\boxed{\begin{minipage}{6cm}$\bigstar$\:\:\sf Median = l + $\sf\dfrac{\frac{n}{2}-C.f.}{f}\times h\\\\Here: \\1)\:n = \sum f =\\2)\:l=Lower\:limit\:of\:median\:class=\\3)\:C.f.=Cumulative\:frequency\:of\:class\\preceeding\:the\:median\:class=\\4)\:f= frequency\:of\:median\:class=\\5)\:h= Class\:interval = \end{minipage}}[/tex]
Answer:
The answer is above this
Explanation:
PLEASE HELP ME 20 POINTS Some metals have a molecular structure that makes them good conductors. Explain how understanding this relationship can help engineers make more powerful batteries.
PLEASE I KNOW THE STRUCTURE I JUST NEED HOW BY UNDERSTANDING THE RELATIONSHIP THEY CAN MAKE MORE POWERFUL BATTERIES.
PLEASE IM DYEING HERE I JUST NEED THIS ANSWER I WILL GIVE BRAINLIST AND 5 STARS A HEART WHATEVER JUST PLEASEEEEE.
Answer:
negative plus positive with energy = battery
Explanation:
Two identical small charged spheres hang in equilibrium with equal masses (0.02kg). The length of the strings is equal (0.18m) and the angle with the vertical is identical (7^o). Find the magnitude of the charge on each sphere. The acceleration of gravity is 9.8 m/s^2 and the value of Coulomb
Answer:
The value is [tex]q = 3.4 *10^{-6} \ C[/tex]
Explanation:
From the question we are told that
The mass of each sphere is [tex]m_1 = m_2 = m = 0.020 \ kg[/tex]
The length of the string is [tex]l = 0.18 \ m[/tex]
The angle of with the vertical is [tex]\theta = 7^o[/tex]
The acceleration due to gravity is [tex]g = 9.8 \ m/s^2[/tex]
Generally the force acting between the forces is mathematically represented as
[tex]F = T cos \theta = \frac{k* q^2}{ r^2}[/tex]
=> [tex]T cos \theta = \frac{k* q^2}{ r^2}[/tex]
Generally from Pythagoras theorem the radius of the circular curve created by the force is
[tex]r = 2 L sin (\theta )[/tex]
=> [tex]r = 2* 0.180 sin (7)[/tex]
=> [tex]r = 0.043 \ m[/tex]
=> [tex]q = tan \theta * \frac{m * g * r^2 }{k}[/tex]
=> [tex]q = tan(7)* \frac{ 0.02 * 9.8 * 0.043^2 }{9*10^{9}}[/tex]
=> [tex]q = 3.4 *10^{-6} \ C[/tex]
A 2.0 kg mass on a horizontal spring is pulled back 2.0 cm and released. If, instead, a 0.40 kg mass were used in this same experiment, the total energy of the system would: ________
a. Remain the same.
b. Double.
c. Be half as large.
Answer:
a. Remain the same.
Explanation:
The total energy of the spring mass system is given as
[tex]$E=\frac{1}{2}kx^2 $[/tex]
where k is the spring constant of the spring
x is the compression or expansion of the spring
The energy of the spring mass system is always independent of mass. It does not depends on the mass object attached to the spring.
Therefore, the total energy of the system will remain the same in the experiment when the mass is changes the second time.
What is the magnitude of the force required to stretch or compress a spring of spring constant 50.0 N/m a distance 2.50 cm from its unstrained length?
a. 50.0 N
b. 125 N
c. 20.0 N
d. 1.25 N
Answer:
Magnitude of the force = 1.25 N
Explanation:
Given:
Spring constant k = 50.0 N/m
Distance s = 2.50 cm = 0.025 m
Find:
Magnitude of the force
Computation:
Magnitude of the force = Ks
Magnitude of the force = 50 x 0.025
Magnitude of the force = 1.25 N
A man is pushing a box of mass m along the ground with a force F. The coefficient of kinetic friction between the box and the ground is μk. The box is accelerating, but then the person stops pushing and the box slides to a halt. The person then starts pushing on the box again with a force F, but the box doesn't budge. The maximum coefficient of static friction between the box and the ground is μs. Identify which of the following statements is true?
a. F > μk/μs
b. μk > μs
c. μkmg < F <= μsmg
d. μsmg= F
Answer: Option C.
Explanation:
When the box is moving, he pushes with a force F.
in this case, the coefficient is μk, and the box is accelerating, this means that the net force is not zero, so F is larger than the friction force.
When the box is still, the man pushes again with a force F, but now the box does not move, so there is no acceleration, which means that the net force is zero, then F is not greater than the maximum static friction force.
Now, the friction force between an object of mass M, and a surface with a coefficient of friction μ is: μ*m*g
where g is the gravitational acceleration.
Then, from the first part, we can conclude that:
μk*m*g < F
(the force F is larger than the kinetic friction force)
and from the second part, we know that:
F ≤ μs*m*g
(The force F is not greater than the static friction force)
If we write those two together, we have:
μk*m*g < F ≤ μs*m*g
Then the correct option is c.
A person travels by car from one city to another with different constant speeds between pairs of cities. She drives for 10.0 min at 75.0 km/h, 6.0 min at 95.0 km/h, and 45.0 min at 40.0 km/h and spends 40.0 min eating lunch and buying gas.(a) Determine the average speed for the trip.
(b) Determine the distance between the initial and final cities along the route.
Answer:
a) v = 0.515 km / min , b) x_total = 52 km
Explanation:
The measured speed is defined by the distance traveled between the time
v = [tex]\frac{\Delta x}{\Delta t}[/tex]
In this case they give us the speed in several time intervals
let's find the distance traveled in each interval
a) Goes at 75 km/h for 10 min
v = [tex]\frac{x}{t}[/tex]x / t
x₁ = v t
let's reduce speed to km / min
v₁ = 75 km / h (1h / 60 min) = 1.25 km / min
the distance traveled in this time is
x₁ = 1.25 10
x₁ = 12.5 km
b) goes to v = 95 km / h for 6 min
v = 95 km / h (1h 60 min) = 1.5833 km / min
the distance traveled is
x₂ = v₂2 t
x₂ = 1.58333 6
x₂ = 9.5 km
c) goes to v = 40 km / h for 45.0 min
v₃ = 40 km / h (1 h / 60min) = 0.6667 km / min
x₃ = 0.6667 45
x₃ = 30 km
d) t = 40 min, stopped
x₄ = 0
A) let's calculate the average speed of the trip
v =[tex]\frac{x_{1}+x_{2}+x_{3}+x_{4} }{t_{1}+t_{2}+t_{3}+t_{4} }[/tex]
v = (12.5 +9.5 +30 +0) / (10 +6 +45 +40)
v = 52/101
v = 0.515 km / min
B) the distance between the two cities is
x_total = x₁ + x₂ + x₃
x_total = 12.5 +9.5 + 30
x_total = 52 km
Which phase of matter does line segment CD represent?
A. plasma
B. liquid
C. gas
D. solid
Answer:
C. Gas
Explanation:
Segment CD represents gaseous phase of matter as segment CD shows highest energy which is in case of gas.
What is matter?
Matter in chemistry, is defined as any kind of substance that has mass and occupies space that means it has volume .Matter is composed up of atoms which may or not be of same type.
Atoms are further made up of sub atomic particles which are the protons ,neutrons and the electrons .The matter can exist in various states such as solids, liquids and gases depending on the conditions of temperature and pressure.
The states of matter are inter convertible into each other by changing the parameters of temperature and pressure.The intermolecular forces of attraction are different in different states of matter.Particles of matter have different sizes.
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The density of mercury is 13.6 g/cm? What is this value in kg/m??
Answer:
68 kg/m
Explanation:
bro
Which of the following is a correct equation for total energy?
Total energy = Kinetic energy / potential energy
Total energy = Kinetic energy - potential energy
Total energy = Kinetic energy * potential energy
Total energy = Kinetic energy + potential energy
Answer:
kinetic energy + potential energy
Answer:
D is the correct answer: Total energy = Kinetic energy + potential energy
You notice electric field lines going towards an isolated charged object in a radial manner. What is the sign of the net charge on the object?a. negative.b. positive.c. neutral.
Answer:
The correct answer is a
Explanation:
The electric field is given by the relation
F = q E
where The force is the Coulomb force and q is a positive test charge, therefore the electric field has the same direction as the electric force.
Consequently if the charge is positive the field must go out of the charge, if the charge is negative the electric field must be directed towards the charge.
Consequently, in this exercise we are told that the lines are directed towards the load, therefore the load must have a negative sign.
The correct answer is a
Light of wavelength 503 nm illuminates a round 0.2-mm diameter hole. A screen is placed 4.9 m behind the slit. What is the diameter of the central bright area on the screen in millimeters
Answer:
30 mm
Explanation:
[tex]\lambda[/tex] = Wavelength of light = 503 nm
d = Diameter of hole = 0.2 mm
D = Distance of screen from light source = 4.9 m
Diameter of the central bright area is given by
[tex]x=\dfrac{2.44\lambda D}{d}\\\Rightarrow x=\dfrac{2.44\times 503\times 10^{-9}\times 4.9}{0.2\times 10^{-3}}\\\Rightarrow x=0.03\ \text{m}=30\ \text{mm}[/tex]
The diameter of the central bright area is 30 mm.
Calculate the magnitude of the normal force on a 15.2 kg block in the following circumstances. (Enter your answers in N.) (a) The block is resting on a level surface. __________ N (b) The block is resting on a surface tilted up at a 35.8 angle with respect to the horizontal. __________ N (c) The block is resting on the floor of an elevator that is accelerating upward at 3.53 m/s ___________ N (d) The block is on a level surface and a force of 155 N is exerted on it at an angle of 35.8 below the horizontal. _________N
Answer:
[tex]149.112\ \text{N}[/tex]
[tex]120.94\ \text{N}[/tex]
[tex]302.77\ \text{N}[/tex]
[tex]58.44\ \text{N}[/tex]
Explanation:
m = Mass of block = 15.2 kg
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
[tex]\theta[/tex] = Angle
a = Acceleration
F = Applied force
Normal force is given by
[tex]N=mg\\\Rightarrow N=15.2\times 9.81\\\Rightarrow N=\boldsymbol{149.112\ \mathbf{N}}[/tex]
[tex]\theta=35^{\circ}[/tex]
[tex]N=mg\cos\theta\\\Rightarrow N=15.2\times 9.81\times \cos35.8^{\circ}\\\Rightarrow N=\mathbf{120.94\ N}[/tex]
[tex]a=3.53\ \text{m/s}^2[/tex]
[tex]N=m(g+a)\\\Rightarrow N=15.2(9.81+3.53)\\\Rightarrow N=\boldsymbol{302.77\ \mathbf{N}}[/tex]
[tex]F=155\ \text{N}[/tex]
[tex]N=mg-F\sin35.8^{\circ}\\\Rightarrow N=15.2\times 9.81-155\sin35.8^{\circ}\\\Rightarrow N=\boldsymbol{58.44\ \mathbf{N}}[/tex]
In a laboratory, it is often convenient to make measurements in centimeters and grams, but SI units are needed for calculations. Convert the following measurements to SI units. (a) 0.92 cm (b) 141.64g kg (c) 15.8 cm3 m3 (d) 63.6 g/cm3 kg/m3
Answer:
(a) 0.92 cm= 0.092 m.
(b) 141.64 g=0.14164 kg.
(c) 15. 8 cm³=0.0000158 m³
(d) 63.6 g/cm³= 63600 kg/m³
Explanation:
The International System of Units, abbreviated S.I., also called the International System of Measurements is a system of measurements in which its units are based on fundamental physical phenomena. The units of the S.I. They are the international reference for the indications of all measuring instruments.
The International System of Units (SI) arose from the need to unify and give coherence to a great variety of unit subsystems.
The International System of Units consists of seven basic units, also called fundamental units, which define the corresponding fundamental physical quantities and which allow any physical quantity to be expressed in terms or as a combination of them. The fundamental physical quantities are complemented by two more physical quantities, called supplementary ones.
By combining the basic units, the other units are obtained, called units derived from the International System, and which allow defining any physical quantity.
(a) The SI unit of length is the meter. Being 1 cm = 0.01 m, then 0.92 cm= 0.092 m.
(b) The SI unit of mass is kg. Being 1 g = 0.001 kg, then 141.64 g=0.14164 kg.
(c) Being 1 cm³ = 0.000001 m³, then 15. 8 cm³=0.0000158 m³
(d) Being 1 g/cm³= 1000 kg/m³, then 63.6 g/cm³= 63600 kg/m³
Does the air in a room or water in a pot have to be at a certain temperature to create convection? Yes or No? and Explain why?
Answer:
Yes
Explanation:
because that temperature creates the bulk movement of molecules due to increase in their kinetic energy
My buddy and i are planning to dive at a site at a site where the water and air temperature are near freezing. We plan to dive to 18 meters for 30 minutes. What dive profile would we use when planning our dive
The dive profile used when planning our dive would be 22 metres/70 feet for 30 minutes.
What is diving?
Diving is the sport or activity of swimming for exploring under water.
My buddy and I are planning to dive at a site where the water and air temperature are near freezing. We plan to dive to 18 meters for 30 minutes.
A General rule as printed on the RDP Table and eRDPML is that if someone planning a dive in cold water that might be harmful. Plan the dive assuming the depth is 4 meters/10 feet deeper than actual. For any other information, see the RDP/eRDPML Instructions for Use booklet.
Thus, dive profile used would be 22 metres/70 feet for 30 minutes.
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Which phenomena result from something other than the flow of thermal energy inside Earth? A-volcanic eruptions B-earthquakes C-thunderstorms D-valley formations
Answer:
D
Explanation:
Valley Formations
1. What is the gravitational force between a 100 kg football player and the Earth at 5.98 x 1024 kg, with a
distance of 6.38 x 106 m?
Answer:
981N
Explanation:
The gravitational force between the football player and the Earth at the given distance is 979.91 N.
The given parameters:
Mass of the player, m = 100 kgMass of the Earth, M = 5.98 x 10²⁴ kgDistance between the bodies, r = 6.38 x 10⁶ mThe gravitational force between the football player and the Earth at the given distance is calculated by applying Newton's law of universal gravitation as follows.
[tex]F = \frac{GmM}{r^2} \\\\F = \frac{6.67 \times 10^{-11} \times 100 \times 5.98 \times 10^{24}}{(6.38 \times ^6)^2} \\\\F = 979.91 \ N[/tex]
Thus, the gravitational force between the football player and the Earth at the given distance is 979.91 N.
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If the elevator car in Figure 9-3A weighs 3,000 newtons, what is the tension in each supporting cable?
Answer:
500 N
Explanation:
An airplane traveling at half the speed of sound emits a sound of frequency 5.84 kHz. (a) At what frequency does a stationary listener hear the sound as the plane approaches?
Answer:
Stationary listener frequency = 77.68 kHz
Explanation:
Given:
Speed of airplane = 344/2 = 172 m/s
Sound of frequency = 5.84 kHz
Computation:
f0 = fs[(v+v0)/(v-vs)]
f0 = 5.84[(344+0)/(344-172)]
f0 = 5.84[(344)/172]
Stationary listener frequency = 77.68 kHz
A car accelerates from rest at 1.0 m/s2 for 20.0 second along a straight road. It then moves at a constant speed for half an hour and then decelerates uniformly to a stop in 30.0 s. Find the total distance covered by the car and the average velocity of the car for the entire trip.
Total distance = 36500 m
The average velocity = 19.73 m/s
Further explanationGiven
vo=initial velocity=0(from rest)
a=acceleration= 1 m/s²
t₁ = 20 s
t₂ = 0.5 hr = 1800 s
t₃= 30 s
Required
Total distance
Solution
State 1 : acceleration
[tex]\tt d=vo.t+\dfrac{1}{2}at^2\\\\d=\dfrac{1}{2}\times 1\times 20^2\rightarrow vo=0\\\\d=200~m[/tex]
[tex]\tt vt=vo+at\\\\vt=at\rightarrow vo=0\\\\vt=1\times 20\\\\vt=20~m/s[/tex]
State 2 : constant speed
[tex]\tt d=v\times t\\\\d=20\times 1800\\\\d=36000~m[/tex]
State 3 : deceleration
[tex]\tt vt=vo+at\rightarrow vt=0(stop)\\\\vo=-at\\\\20=-a.30~s\\\\a=-\dfrac{2}{3}m/s^2(negative=deceleration)[/tex]
[tex]\tt d=vot+\dfrac{1}{2}at^2\\\\d=20.30-\dfrac{1}{2}.\dfrac{2}{3}.30^2\\\\d=300~m[/tex]
Total distance : state 1+ state 2+state 3
[tex]\tt 200 + 36000 + 300=36500~m[/tex]
the average velocity = total distance : total time
[tex]\tt avg~velocity=\dfrac{36500}{20~s+1800~s+30~s}=19.73~m/s[/tex]
The total distance covered by the car will be 36500m and the average velocity of the car for the entire trip will be 19.73 m/sec.
What is average velocity?
The average velocity is the ratio of the total distance traveled to the total time taken by the body. Its unit is m/sec.
The given data in the problem is;
v₀ is the initial velocity=0(from rest)
a is the acceleration= 1 m/s²
t₁ is the time period 1= 20 s
t₂ is the time period 2= 0.5 hr = 1800 s
t₃ is the time period 3= 30 s
For part 1
The distance travelled is found as;
[tex]\rm d=v_0t+\frac{1}{2} at^2 \\\\ \rm d=\frac{1}{2}\times 1 \times (20)^2 \\\\ \rm d=200\ m[/tex]
The initial velocity is found as ;
[tex]\rm v_t=v_0+at\\\\ \rm v_t=0+1 \times 20\\\\ \rm v_t=20 \ m/sec[/tex]
For part 2
The distance traveled is found as;
[tex]\rm d= v \times t \\\\ \rm d= 20 \times 1800 \\\\ \rm d= 36000 \ m[/tex]
For part 3
The acceleration is found as;
[tex]\rm v_t=v_0+at\\\\ \rm 0=v_0+a \times 30\\\\ \rm 20=-30 a \\\\ a= - \frac{2}{3}\ m/sec^2[/tex]
The distance traveled is found as;
[tex]\rm d=v_0t+\frac{1}{2} at^2 \\\\ \rm d=20.30 +\frac{1}{2} \times \frac{2}{3} (30)^2 \\\\ \rm d=300 \ m[/tex]
Total distance travelled= distance travelled( part 1+ part 2+ part3)
Total distance travelled=200+3600+300 = 36500 m
The average velocity is found as;
[tex]v_{avg}= \frac{36500}{20+1800+30} \\\\ v_{avg}= 19.73 \ sec[/tex]
Hence the total distance covered by the car will be 36500m and the average velocity of the car for the entire trip will be 19.73 m/sec.
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Jane is riding in a hot air balloon that is rising vertically at a constant speed of 3 m/s over a lake. She drops a rock from the balloon when the distance from the rock to the water is 50 m. Use g m/s^2, and let the up direction be positive. How long after Jane drops the rock will it splash into the water?
It is given that up direction be positive.
Ball velocity at the time is was dropped is 3 m/s.
g = -9.8 m/s²
s = -50 m
Now, using equation of motion for time :
s = ut + at²/2
3t - 4.9t² + 50 = 0
4.9t² - 3t - 50 = 0
Solving above equation, we get :
t = 3.52 s
Therefore, the rock will splash into the water after 3.52 s.
(a) A defibrillator passes 12.0 A of current through the torso of a person for 0.0100 s. How much charge moves
Answer:
0.12C
Explanation:
A defibrillator passes 12A through the torso of an individual
The time is 0.0100
The charge can be calculated as follows
= I × t
= 12 × 0.0100
= 0.12C
Hence the charge is 0.12C
A neutron has
A. an alternating charge.
B. a negative charge.
C. a positive charge.
D. a neutral charge.
Answer:
D
Explanation:
a neutron does not have a positive nor negative charge it remains neutral
How much force does it take to accelerate a car mass 1750kg at 20m/s
Answer:
The needed force is 35,000 N.
Explanation:
By the second Newton's law, we know that:
F = M*a
Force equals mass times acceleration.
If we know that the mass of the car is 1750kg, then M = 1750kg, and the acceleration is 20 m/s^2, then a = 20m/s^2
We can replace these values in the original equation to find the value of F.
F = 1750kg*20 m/s^2 = 35,000 N
A certain part of the electromagnetic spectrum ranges from 200 nm to 400 nm. What is the highest frequency associated with this portion of the spectrum
Answer:
1.5* 10¹⁵ Hz
Explanation:
In any electromagnetic wave (assuming it's propagating in vacuum), the propagation speed is just c, the speed of light.As in any wave, there exists a fixed relationship between speed (c), frequency (f) and wavelength (λ), as follows:[tex]c = \lambda * f (1)[/tex]
Since c is an universal constant, we can see that there exists an inverse relationship between λ and f, so if the lower the wavelength, the higher the frequency.So, in this case, the highest frequency will be the one associated with the lowest wavelength, i.e., 200 nm.Solving for f from (1), replacing by the values, we get:[tex]f = \frac{c}{\lambda} = \frac{3e8m/s}{200e-9m} = 1.5e15 1/sec = 1.5e15 Hz (2)[/tex]